I tried to solve it using the Euclidean algorithm but after finding the gcd$(15, 28)=$ $1$, and reversing the steps of the Euclidean algorithm I get the following expression:
These are the reversal steps: $$13=2\times6+1$$ $$1=13-(2\times6)$$ $$1=13-6\left(15-13\right)$$ $$1=28-15-6\left(15-13\right)$$ $$1=28-15\left(1+6\right)+6\text{·}13$$
The expected result should be some expression of the form $$1=28\left(y\right)-15\left(x\right)$$
On
You already replaced a 13 by $$13=28-15$$
Why don't you replace the last 13 also by this?
It would have been faster if you combined the 13's together before replacing.
On
The fast method uses the extended Euclidean algorithm, based on the observation that all reminders in the Euclidean algorithm can be written as a linear combination of $15$ and $28$, with a recursive relation deduced from the relation step $i$: $$ r_{i-1}=q_i r_i+r_{i+1}\iff r_{i+1}= r_{i-1}-q_i r_i,$$ from which we deduce the relations (with obvious notations) $$ x_{i+1}= x_{i-1}-q_i x_i,\qquad y_{i+1}= y_{i-1}-q_i y_i.$$ As it is a recursion of order $2$, we have to set the initial values $r_0=15$, $r_{-1}=28$, whence the following layout:
\begin{array}{rrrl} r_i &x_i&y_i&q_i \\ \hline 28 & 0 & 1 \\ 15 & 1 & 0 & 1 \\ \hline 13 & -1 & 1 & 1 \\ 2 & 2 & -1 & 6 \\ \color{red}1 & \color{red}{-13} & \color{red}{7} \\ \hline \end{array}
On
Applying the Euclidean algorithm we have $$28=1\cdot15+13$$ $$15=1\cdot13+2$$ $$13=6\cdot2+1$$ $$2=2\cdot1+0$$
So we have $\gcd(13,28)=1$ and for the reversal steps $$1=13-6\cdot2$$ and now substituting the second $$2=15-1\cdot 13$$ we have $$1=13-6\cdot2=13-6\cdot(15-1\cdot13)=7\cdot13-6\cdot15$$ and finally substituting the first $13=28-1\cdot15$ we obtain $$1=7\cdot13-6\cdot15=7\cdot(28-1\cdot15)-6\cdot15$$ $$=7\cdot28-13\cdot15$$
Thus $$7\cdot28-13\cdot15=1$$
$1=13-6\times2=13-6(15-13)=7\times13-6\times15=7\times(28-15)-6\times15$
$=7\times28-13\times15$